在 C++ 中传递具有移动语义的 unique_ptr 向量

Question

I am learning CPP++14 move semantics.While writing a small code I observed some weird behavior. I am moving vector of unique ptr to a function using r-value refrence. on debuuging I found that the changes are being applied to the moved object also. Why am I observing this hcnage even the object is moved? Whats does the move do in following code?

void func(std::vector<std::unique_ptr<int>> && vect) {
    vect.emplace_back(std::move(std::make_unique<int>(3)));
    return ;
}

int  main() {
    std::vector<std::unique_ptr<int>> a;
    func(std::move(a));
    cout<<(*(a[0]))<<endl;
    return 0;
}

Answer 1

Whats does the move do in following code?

Move operation is not performed in func(std::move(a)); in fact, std::move just performs conversion and produces an rvalue (xvalue) expression, which is just bound to the rvalue reference parameter vect of func . Then any modification on vect inside func has effect on the argument (ie a ) too, they refer to the same object.

In particular, std::move produces an xvalue expression that identifies its argument t . It is exactly equivalent to a static_cast to an rvalue reference type.

If you change the parameter to pass-by-value, then you'll see move operation is performed. And given the usage you showed, just pass-by-lvalue-reference seems less confusing (and no need to use std::move on argument again).

BTW: In vect.emplace_back(std::move(std::make_unique<int>(3))); the usage of std::move is superfluous, std::make_unique<int>(3) been an rvalue expression.